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marissa [1.9K]
3 years ago
5

Prove that tan60+tan70-tan50+tan10=2sqrt3?

Mathematics
1 answer:
LiRa [457]3 years ago
8 0
{tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)} 

<span>ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to: </span>

<span>= 8*tan(10)/{1 - 3*tan²(10)} </span>

<span>iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10) </span>

<span>= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= 3*tan(30) = 3*(1/√3) = √3 [Proved] </span>

<span>[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)}, </span>
<span>{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]</span>
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A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic
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Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean iron concentration = 0.802 cc/cubic meter

             s = sample standard deviation = 0.093

             n = number of water samples = 27

             \mu = population mean

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, \mu is ;

P(-1.706 < t_2_6 < 1.706) = 0.90  {As the critical value of t at 26 degree of

                                                 freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 1.706) = 0.90

P( -1.706 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 1.706 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X -1.706 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +1.706 \times {\frac{s}{\sqrt{n} } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X -1.706 \times {\frac{s}{\sqrt{n} } , \bar X +1.706 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 0.802 -1.706 \times {\frac{0.093}{\sqrt{27} } , 0.802 +1.706 \times {\frac{0.093}{\sqrt{27} } ]

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Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

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Step-by-step explanation:

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Step-by-step explanation:

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From the diagram attached

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